fit a curve to data without using curve fitting toolbox
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Anqi Li
el 12 de Dic. de 2014
Comentada: Andreas Goser
el 12 de Dic. de 2014
I have a set of data=[x1 x2] which looks periodical. I want to fit them into this Fourier transform equation:
x2 = A1 + A2.*sin(x1) + A3.*cos(x1) + A4.*sin(2*x1) + A5.*cos(2*x1)
by using least squares optimisation to know the optimum A = [A1;A2;A3;A4;A5]
I don't have curve fitting toolbox. please let me know how to do it without using toolbox.
1 comentario
Andreas Goser
el 12 de Dic. de 2014
I see you already have answers, but I wonder why you do not have the Curve Fitting Toolbox. Is the information about the university you work at current (your profile)?
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Mohammad Abouali
el 12 de Dic. de 2014
Editada: Mohammad Abouali
el 12 de Dic. de 2014
Assuming your X1 and X2 are vector, i.e. size(x1)=size(x2)=[N 1] and N>=5 (since there are 5 coefficients, A1 to A5); then
C=[ones(numel(x1),1) sin(x1(:)) cos(x1(:)) sin(2*x1(:)) cos(2*x1(:))];
A=(C'*C)\(C'*x2(:));
or even:
A=C\x2(:);
6 comentarios
Mohammad Abouali
el 12 de Dic. de 2014
so x2(:) makes sure that your data is a column vector data, i.e. size(x2(:))=[N 1]. I just wanted to be sure that it is not a row data but a vertical column vector. If it is already a column vector you can just replace it with X2 and drop (:).
A=C\x2 conceptually is pretty much solving C*A=x2; Your system of equations can be written as matrix C (the one with sine and cosine functions) multiplied by the unknown column vector, and x2 is your known variable vectors. once you do A=C\x2 pretty much you are solving the system of linear-equations. This operation is known as mldivide. If you want to know more about it go to mldivide help section, there you can find more information on how to use this.
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